If sum of first 6 terms of an AP is 36

Solution:

Let a and d be the first term and common difference, respectively of an AP

$\because$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\ldots$ (i)

Now, $S_{6}=36$ [given]

$\Rightarrow \quad \frac{6}{2}[2 a+(6-1) d]=36$

$\Rightarrow$ $2 a+5 d=12$ (ii)

$\Rightarrow \quad \frac{16}{2}[2 a+(16-1) d]=256$

$\Rightarrow \quad 2 a+15 d=32$ ...(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

$10 d=20 \Rightarrow d=2$

From Eq. (ii), $2 a+5(2)=12$        $2 a+5(2)=12$

$\Rightarrow \quad 2 a=12-10=2$

$\Rightarrow \quad a=1$

$\therefore$ $S_{10}=\frac{10}{2}[2 a+(10-1) d]$

$=5[2(1)+9(2)]=5(2+18)$

 

$=5 \times 20=100$

Hence, the required sum of first 10 terms is 100.